Problem 9 - Thomas Calculus

Question

Evaluate \(\int _ { C } ( x + y ) d s\) where \(C\) is the straight-line segment \(x = t , y = ( 1 - t ) , z = 0 ,\) from \(( 0,1,0 )\) to \(( 1,0,0 )\)

Step-by-Step Solution

Answer

The integral evaluates to \(\sqrt{2}\).

Step 1: Parameterize the Curve C

The given curve segment from \((0,1,0)\) to \((1,0,0)\) is represented by the parametric equations \(x = t, \, y = (1 - t), \, z = 0\) where \(t\) ranges from \(0\) to \(1\). This describes a straight line path between the given points.

Step 2: Find the Expression for ds

To find \(ds\), compute the differential arc length as follows:\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Compute the derivatives \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = -1\). Then,\[ ds = \sqrt{1^2 + (-1)^2} \, dt = \sqrt{2} \, dt \]

Step 3: Express the Integrand (x + y)

Given \((x + y)\), substitute the parametric expressions:\[ x + y = t + (1 - t) = 1 \]Notice that \(x + y\) simplifies to 1 over the entire range of \(t\).

Step 4: Set Up the Integral

The line integral \(\int_C (x+y) \, ds\) transforms into:\[ \int_0^1 1 \times \sqrt{2} \, dt \]

Step 5: Solve the Integral

Evaluate the integral:\[ \int_0^1 \sqrt{2} \, dt = \sqrt{2} \left[ t \right]_0^1 = \sqrt{2} (1 - 0) = \sqrt{2} \]

Key Concepts

Parametric Equations

Parametric equations are a set of equations that express the coordinates of the points that make up a geometric object as functions of a variable, usually denoted by \( t \). In the context of line integrals, parametric equations are used to describe the path or curve along which integration is performed.
For the given exercise, the line segment from \((0, 1, 0)\) to \((1, 0, 0)\) is characterized by the parametric equations:

• \( x = t \)
• \( y = 1 - t \)
• \( z = 0 \)

Here, \( t \) represents the parameter, and its range from 0 to 1 describes the entire line segment. This straightforward setup is ideal for evaluating the integral along a straight path.

Differential Arc Length

The differential arc length \( ds \) is a small segment of length along the curve. It provides a measure of distance that can be used in line integrals. The formula to compute \( ds \) in terms of parametric equations is given by:
\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]This formula derives from the Pythagorean theorem, extended to tiny linear segments along the curve.

• In the example, compute derivatives: \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = -1 \).
• Substituting these into the formula gives: \( ds = \sqrt{1^2 + (-1)^2} \, dt = \sqrt{2} \, dt \).

This calculation simplifies the concept of distance for the line segment, making integration feasible.

Integral Evaluation

Once parametric equations and \( ds \) are defined, the next step is integration. This involves setting up and evaluating the integral over the parameter's range.
The integrand in the exercise is \( x + y \), which is simplified using the parametric forms:

• \( x + y = t + (1 - t) = 1 \)

The simplification often eases computations, especially if it results in a constant, as in this example. Consequently, the integral becomes:
\[ \int_C (x+y) \, ds = \int_0^1 1 \times \sqrt{2} \, dt \]This is a straightforward evaluation since the integrand is constant over the domain of \( t \).

• Calculate: \( \int_0^1 \sqrt{2} \, dt = \sqrt{2} \times (1 - 0) = \sqrt{2} \)

This final evaluation completes the line integral along the curve.

Vector Calculus

Vector calculus is a branch of mathematics that extends calculus to vector-valued functions. It is often employed to analyze and compute integrals over curves and surfaces using vectors.
In line integrals like the one in the exercise, vector calculus allows expressing curves with vector functions. The parametric equations themselves can be framed in terms of vectors:

• The position vector for the line segment can be represented as \( \mathbf{r}(t) = \langle t, 1-t, 0 \rangle \).

Vector calculus also aids in understanding the direction and magnitude of curves through derivatives and integrals.
It helps to correlate various aspects, like setting up the differential arc length \( ds \) using derivatives, which are inherently vector operations. Combining vectors with calculus gives a powerful tool for handling complex integrals in multidimensional spaces.